Derivative of an expoential within an exponential

[sprint]

[SOLVED] derivative of an expoential within an exponential

help. i need help on finding the derivative of an exponential within an exponential

Homework Statement

d/dx of e^(e^4x)

Homework Equations

d/dx of e^(e^4x)

The Attempt at a Solution

d/dx of e^(e^4x)

i don't know how to attempt this cause the function I am interested is in the power of e

 

[hotcommodity]

We know that the derivative of an exponential is simply the exponential times the derivative of power term, right?

\frac{d}{du} e^u = e^uu'

You're going to have to apply the chain rule.

Does that help?

 

[sprint]

We know that the derivative of an exponential is simply the exponential times the derivative of power term, right?

\frac{d}{du} e^u = e^uu'

You're going to have to apply the chain rule.

Does that help?

i think we are not looking at the same problem. my problem is an exponential within an exponential

d/dx e^(e^4x)

 

[hotcommodity]

It's not the exact equation that you need to apply, but that's where you start. If I let u = e^{4x}, then I must first find the derivative of e^u.

I know \frac{d}{du} e^u = e^uu'.

So you know what "u" is, now you must find the derivative of "u" and plug it into the above equation.

 

[sprint]

oooooh.

i see. i guess just use substitution letting u = e^(4x)

by any chance, is the correct answer 4e^(4x) times e^(4x) = 4e^(8x)

?

 

[hotcommodity]

Not quite, you want to think of "u" as separate from everything else and only plug it in at the end. If u = e^{4x}, what's the derivative, u' ?

 

[sprint]

thats exactly what i did. i made "u" separate.

and the derivative of u is 4e^(4x), to answer your question.

 

[hotcommodity]

Right, so you'd have 4e^{e^{4x}}e^{4x}. I didn't see you type in the e^{e^{4x}} part.

 

[sprint]

wow. totally confusing.

but i think the answer is what i stated earlier...

4e^(4x) times e^(4x)

i get this by sticking to the basics and using the chain rule and substitution

 

[hotcommodity]

I'm not trying to confuse you, haha, but I'd hope that you walk away from this understanding what's going on. Let's say I wanted to find the derivative of e^{e^x}. I'd use the chain rule once again letting u = e^x. So I'd have:

\frac{d}{du} e^u = e^uu'

Find u':

u' = e^x * 1

And plug u and u' into the first equation to get the derivative:

e^{e^x}*e^x*1

Right?

 

[sprint]

check it. i was wrong. the answer should be

4e^(4x) times e^(e^(4x))

which is

u' times e^u

via the chainrule

 

[sprint]

I'm not trying to confuse you, haha, but I'd hope that you walk away from this understanding what's going on. Let's say I wanted to find the derivative of e^{e^x}. I'd use the chain rule once again letting u = e^x. So I'd have:

\frac{d}{du} e^u = e^uu'

Find u':

u' = e^x * 1

And plug u and u' into the first equation to get the derivative:

e^{e^x}*e^x*1

Right?

i follow you. you are right.

 

[sprint]

and how do you type in the equations like that? it looks clean...

 

[hotcommodity]

Haha, you just use the "tex" tags. You can click on the equation itself to see how you would type it in. I think there's a tutorial somewhere here on the site.

 

[sprint]

ok. thanks. lol.

 

[hotcommodity]

You're quite welcome :)